Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 1402

The axis of the parabola \(9 \mathrm{y}^{2}-16 \mathrm{x}-12 \mathrm{y}-57=0\) is (a) \(\mathrm{y}=0\) (b) \(16 \mathrm{x}+61=0\) (c) \(3 \mathrm{y}-2=0\) (d) \(3 \mathrm{y}-61=0\)

Short Answer

Expert verified
The axis of the given parabola is \[\boxed{16x + 61 = 0}\].

Step by step solution

01

Rewrite the given equation into standard form

First, we need to rewrite the given equation \(9y^2-16x-12y-57=0\) into the standard form of a parabolic equation. The standard form of a parabolic equation is either: \(y - k = a(x - h)^2\) (a vertical parabola) or \(x - h = a(y - k)^2\) (a horizontal parabola) Where (h, k) is the vertex of the parabola and a is a constant that controls the shape of the parabola. Rearrange the given equation as: \(9y^2 - 12y = 16x + 57\) Now complete the square for the y terms: \((y - \frac{b}{2a})^2\), where in this case, a = 9 and b = -12.
02

Completing the Square and Rearranging

To complete the square for the y terms, we need to add and subtract \(\left(\frac{-12}{2 \cdot 9}\right)^{2}\). \((y - \frac{-12}{2 \cdot 9})^2 = (-\frac{2}{3})^2 = \frac{4}{9}\) Now rewrite the equation, completing the square for y: \(9\left(y^2 -\frac{4}{3}y + \frac{4}{9}\right) = 16x + 57 + 9\left(\frac{4}{9}\right)\) Then, simplify the equation: \(9 \left(y - \frac{2}{3}\right)^2 = 16x + 61\) Now, we have the standard form of a parabolic equation: \(x - h = a(y - k)^2\) Comparing the given equation with standard form: \(x - (-61/16) = \frac{1}{16} \left(y-\frac{2}{3}\right)^2\)
03

Identify the Axis of Parabola

Our equation is of the form \(x - h = a(y - k)^2\), which represents a horizontal parabola. The axis of this parabola is parallel to the y-axis. From the standard equation, we can see that the axis is a vertical line with the equation: \(x = -\frac{61}{16}\) Comparing this to the given options, it matches option (b), which is: \(16x + 61 = 0\) Therefore, the axis of the given parabola is: \[\boxed{16x + 61 = 0}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{ax}+\mathrm{cy}+\mathrm{a}=0\) and \(\mathrm{x}^{2}+\mathrm{y}^{2}-3 \mathrm{ax}+\mathrm{dy}-1=0\) intersect in two distinct points \(\mathrm{P}\) and \(Q\), then the line \(5 x+\) by \(-a=0\) passes through \(P\) and \(Q\) fore (a) no value of a (b) exactly one value of a (c) exactly two values of a (d) infinitely many value of a

The shortest distance between the line \(\mathrm{x}-\mathrm{y}+1=0\) and the curve \(\mathrm{x}=\mathrm{y}^{2}\) is \(\ldots \ldots \ldots\) (a) \([(3 \sqrt{2}) / 5]\) (b) \([(2 \sqrt{3}) / 8]\) (c) \([(3 \sqrt{2}) / 8]\) (d) \([(2 \sqrt{2}) / 5]\)

Let \(E\) be the ellipse \(\left(x^{2} / 9\right)+\left(y^{2} / 4\right)=1\) and \(C\) be the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}=9\). Let \(\mathrm{P}\) and \(\mathrm{Q}\) be the point \((1,2)\) and \((2,1)\) respe. Then (a) P lies inside \(\mathrm{C}\) but outside \(\mathrm{E}\) (b) P lies inside both \(\mathrm{C}\) and \(\mathrm{E}\) (c) \(Q\) lies outside both \(\mathrm{C}\) and \(\mathrm{E}\) (d) Q lies inside \(\mathrm{C}\) but outside \(\mathrm{E}\)

The length of the chord joining the points \((2 \cos \theta, 2 \sin \theta)\) and \(\left(2 \cos \left(\theta+60^{\circ}\right), 2 \sin \left(\theta+60^{\circ}\right)\right)\) of the circle \(x^{2}+y^{2}=4\) is (a) 2 (b) 4 (c) 8 (d) 16

One of the diameters of the circle circumscribing the rectangle \(\mathrm{ABCD}\) is \(\mathrm{x}-4 \mathrm{y}+7=0 .\) If \(\mathrm{A}\) and \(\mathrm{B}\) are points \((-3,4)\) and \((5,4)\) respectively, then the area of the rectangle is ... (a) 32 sq. units (b) 16 sq. units (c) 64 sq. units (d) 8 sq. units
See all solutions

Recommended explanations on English Textbooks

Language and Social Groups

Read Explanation

Phonology

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Free Response Essay

Read Explanation

Linguistic Terms

Read Explanation

Semiotics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free